Hi!!! i am using Matlab for the first time!) Please, Can u Help me to solve a non-linear system?!

Hi!!! i am using Matlab for the first time!) Please, Can u Help me to solve a non-linear system?! i have seen some videos about it,, then did what i had seen) but there is an error: <Strings passed to EVAL cannot contain function declarations>, but i don't know what is wrong.. Pleeease, help!! here is the system:

fcns(1)=-y*(2*y^2+2*x^2+4*x^2*y^2+2*x*y-4*x*y^2-4*x^2*y)^(1/2)/(x+y-2*x*y)+1/2*(1-x*y)/(2*y^2+2*x^2+4*x^2*y^2+2*x*y-4*x*y^2-4*x^2*y)^(1/2)/(x+y-2*x*y)*(4*x+8*x*y^2+2*y-4*y^2-8*x*y)-(1-x*y)*(2*y^2+2*x^2+4*x^2*y^2+2*x*y-4*x*y^2-4*x^2*y)^(1/2)/(x+y-2*x*y)^2*(1-2*y);

AND

fcns(2)= -x*(2*y^2+2*x^2+4*x^2*y^2+2*x*y-4*x*y^2-4*x^2*y)^(1/2)/(x+y-2*x*y)+1/2*(1-x*y)/(2*y^2+2*x^2+4*x^2*y^2+2*x*y-4*x*y^2-4*x^2*y)^(1/2)/(x+y-2*x*y)*(4*y+8*x^2*y+2*x-8*x*y-4*x^2)-(1-x*y)*(2*y^2+2*x^2+4*x^2*y^2+2*x*y-4*x*y^2-4*x^2*y)^(1/2)/(x+y-2*x*y)^2*(1-2*x);

[bym]

what do you mean by 'solve'... is fcns(1) = fcns(2)?

[Aizhan Merey]

no,,)) I meant that fcns(1)=0; fcns(2)=0 :)

Answer 1

[Walter Roberson]

The equations are symmetric: if you exchange the variables x and y in the first equation, you get the second equation. Therefore at least one of the solution (probably all of the real solutions) fall on to x=y. So substitute y=x in the first equation and simplify and solve that, and you will get at least one of the solutions (probably the only real one.)

My tests using solve() indicate that there is in fact one real solution, and 4 closely related imaginary solutions (which become obvious if you factor the equations and examine the numerator.)

[bym]

@walter - whoops, sorry I didn't see your post which is much more helpful than mine +1 vote (not that you need it)

[Aizhan Merey]

THANK YOU A LOOOOOOT!!!!! :D,,

Answer 2

[bym]

Using the symbolic toolbox, I got the following using solve() which assumes both expressions are equal to 0. note that i is imaginary symbol (sqrt(-1))

a.x

ans =

         1/2
           0
   i/2 + 1/2
           0
   1/2 - i/2
  a.y

ans =

         1/2
   i/2 + 1/2
           0
   1/2 - i/2
           0

[Aizhan Merey]

THANK YOU A LOOOOOOT!!!!! :D,,

[Aizhan Merey]

eem.. wait,, sqrt(-1)??? oO is it possible?? u mean that i= sqrt(-1)?? x) sorry,, didn't understand..

[Walter Roberson]

Yes, the i/2 in the answers means sqrt(-1)/2 . As I mentioned above, there is one solution that is completely real-valued, and four solutions that are complex.